The infinite sequence of prime numbers

[Gribouillis]

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#!/usr/bin/env python # SPDX-FileCopyrightText: 2022 Member Gribouillis of www.python-forum.io # # SPDX-License-Identifier: MIT   __version__ = '2022.01.28.2'   from heapq import heappush, heappop, heappushpop import itertools as itt     def primes():     """Return the infinite sequence of all prime numbers"""     yield 2     h = []     n = 3     x, i, seq = 4, 2, itt.count(6, 2)     while True:         if n < x:             yield n             n2 = n ** 2             heappush(h, (n2, n, itt.count(n2 + n, n)))         n = x + 1         x, i, seq = heappushpop(h, (next(seq), i, seq))   if __name__ == '__main__':           s = primes()     # print the first prime numbers     for i in range(100):         p = next(s)         print(p)

[lucasbazan]

I made a function that verify if a number is prime, hope that is can help your code to be more tiny

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def is_prime(num):     mult = [i for i in range(1, num+1) if num % i == 0]     if len(mult) == 2 and 1 in mult and num in mult:         return True     return False

https://gist.github.com/lucasbazan/9a3c4...3431c55532

[Gribouillis]

@lucasbazan How can it help?

[lucasbazan]

@lucasbazan How can it help?

If I understand correctly you can create an infinite loop with a counter using this function of mine, so I believe your code can be short. Some thing like this:

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def is_prime(num):     mult = [i for i in range(1, num+1) if num % i == 0]     if len(mult) == 2 and 1 in mult and num in mult:         return True     return False   def infinite_primes():     counter = 1     while True:         if is_prime(counter):             print(counter)         counter += 1   if __name__ == '__main__':     infinite_primes()

[Gribouillis]

The code is short but the performance is terrible. I modified your code to compare the two functions. Here are the results

Output:
primes(): 0.0711168740017456 seconds
infinite_primes(): 55.40710521499568 seconds
infinite_primes() is 779 times slower on 5000 primes.

The results rapidly get worse as the number of primes increase...

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from heapq import heappush, heappop, heappushpop import itertools as itt       def primes():     """Return the infinite sequence of all prime numbers"""     yield 2     h = []     n = 3     x, i, seq = 4, 2, itt.count(6, 2)     while True:         if n < x:             yield n             n2 = n ** 2             heappush(h, (n2, n, itt.count(n2 + n, n)))         n = x + 1         x, i, seq = heappushpop(h, (next(seq), i, seq))      def is_prime(num):     mult = [i for i in range(1, num+1) if num % i == 0]     if len(mult) == 2 and 1 in mult and num in mult:         return True     return False    def infinite_primes():     counter = 1     while True:         if is_prime(counter):             yield counter         counter += 1   if __name__ == '__main__':     from collections import deque     from time import perf_counter     exhaust = deque(maxlen=0).extend           def perf(seq, n):         start = perf_counter()         exhaust(itt.islice(seq, 0, n))         return perf_counter() - start       N = 5000     t1 = perf(primes(), N)     print(f'primes(): {t1} seconds')     t2 = perf(infinite_primes(), N)     print(f'infinite_primes(): {t2} seconds')     print(f'infinite_primes() is {int(t2/t1)} times slower on {N} primes.')